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A 0.950 kgkg block is attached to a spring with spring constant 15 N/mN/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 47 cm/scm/s .What is the amplitude of the subsequent oscillations? Answer should be in cm.

Respuesta :

Answer:

Explanation:

Initial compression in the spring due to hit by hammer will create amplitude of oscillation.

If x be the compression  in the spring after hit

1/2 k x² = 1/2 m v²

15 x² = .95 x (47 x 10⁻²)²

x² = 140 x 10⁻⁴

x = 11.83 x 10⁻² m

11. 83 cm  

Amplitude of oscillation = 11.83 cm

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